NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
October 24, 2020, 10:33:59 am

There is a better way to do it; find it. ..."Thomas Edison(1847-1931, American inventor, 1093 patients)"

 Pages: [1]   Go Down
 Author Topic: work  (Read 21530 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Guest
 1 work « Embed this message on: February 06, 2004, 03:29:00 am »

A rectangulat tank that is three feet long and two feet wide and two feet deep contains water that is 1.5 feet in depth calculate the amount of work required to pump half of the water over the top of the tank (weight density of water is 62.5 pounds per cubic foot)?
 Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3086

 « Embed this message Reply #1 on: February 06, 2004, 04:41:26 pm » posted from:Taipei,T'ai-pei,Taiwan

Use conservation of energy:
Calculate the total potential energy (related to water surface) for upper half of water !
$U= \int dm \cdot g \cdot y =\int \rho \cdot dV \cdot g \cdot y =\int \rho \cdot A \cdot dy \cdot g \cdot y$
where $\rho$ is the density, A is the area
 Logged
Newbie

Offline

Posts: 3

 « Embed this message Reply #2 on: June 29, 2007, 01:39:38 pm »

A little bit of confused... What is that?
 Logged
 Pages: [1]   Go Up
There is a better way to do it; find it. ..."Thomas Edison(1847-1931, American inventor, 1093 patients)"